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数列求和

\[ 1 + 2 + \cdots + n = \frac{1}{2}n\left( {n + 1} \right) \] \[ 1^2 + 2^2 + 3^2 + \cdots + n^2 = \frac{1}{6}n\left( {n + 1} \right)\left( {2n + 1} \right) \] \[ 1^3 + 2^3 + 3^3 + \cdots + n^3 = \frac{1}{4}n^2 \left( {n + 1} \right)^2 \]

\[ 1^4 + 2^4 + 3^4 + \cdots + n^4 \] \[= \frac{1}{{30}}n\left( {n + 1} \right)\left( {2n + 1} \right)\left( {3n^2 + 3n - 1} \right) \]

\[ 1^5 + 2^5 + 3^5 + \cdots + n^5 \] \[= \frac{1}{{12}}n^2 \left( {n + 1} \right)^2 \left( {2n^2 + 2n - 1} \right) \]

\[ 1^6 + 2^6 + 3^6 + \cdots + n^6 \] \[= \frac{1}{{42}}n\left( {n + 1} \right)\left( {2n + 1} \right)\left( {3n^4 + 6n^3 - 3n + 1} \right) \]

\[ 1^7 + 2^7 + 3^7 + \cdots + n^7 \] \[= \frac{1}{{24}}n^2 \left( {n + 1} \right)^2 \left( {3n^4 + 6n^3 - n^2 - 4n + 2} \right) \]

\[ 1 - 2 + 3 - \cdots + \left( { - 1} \right)^{n - 1} n = \left\{ {\begin{array}{*{20}c} {\frac{1}{2}\left( {n + 1} \right)n,n为奇数} \\ { - \frac{n}{2},n为偶数} \\ \end{array}} \right. \]

\[ 1^2 - 2^2 + 3^2 - \cdots + \left( { - 1} \right)^{n - 1} n^2 = \left( { - 1} \right)^{n - 1} \frac{1}{2}n\left( {n + 1} \right) \]

\[ 1^3 - 2^3 + 3^3 - \cdots + \left( { - 1} \right)^{n - 1} n^3= \left\{ {\begin{array}{*{20}c} {\frac{1}{4}\left( {2n - 1} \right)\left( {n + 1} \right)^2 ,n为奇数} \\ { - \frac{1}{4}n^2 \left( {2n + 3} \right),n为偶数} \\ \end{array}} \right. \]

\[ 1^4 - 2^4 + 3^4 - \cdots + \left( { - 1} \right)^{n - 1} n^4 = \left( { - 1} \right)^{n - 1} \frac{1}{2}n\left( {n + 1} \right)\left( {n^2 + n - 1} \right) \]

\[2 + 4 + 6 +  \cdots  + 2n = n\left( {n + 1} \right)\]

\[ 1 + 3 + 5 + \cdots + \left( {2n - 1} \right) = n^2 \]

\[ 1^2 + 3^2 + 5^2 + \cdots + \left( {2n - 1} \right)^2 = \frac{1}{3}n\left( {4n^2 - 1} \right) \]

\[ 1^3 + 3^3 + 5^3 + \cdots + \left( {2n - 1} \right)^3 = n^2 \left( {2n^2 - 1} \right) \]

\[ 1 \cdot 2 + 2 \cdot 3 + 3 \cdot 4 + \cdots + n\left( {n + 1} \right) = \frac{1}{3}n\left( {n + 1} \right)\left( {n + 2} \right) \]

\[ 1 \cdot 2 \cdot 3 + 2 \cdot 3 \cdot 4 + 3 \cdot 4 \cdot 5 + \cdots + n\left( {n + 1} \right)\left( {n + 2} \right) \] \[ = \frac{1}{4}n\left( {n + 1} \right)\left( {n + 2} \right)\left( {n + 3} \right) \]

\[ 1 \cdot 2 \cdot 3 \cdot 4 + 2 \cdot 3 \cdot 4 \cdot 5 + \cdots + n\left( {n + 1} \right)\left( {n + 2} \right)\left( {n + 3} \right) \] \[ = \frac{1}{5}n\left( {n + 1} \right)\left( {n + 2} \right)\left( {n + 3} \right)\left( {n + 4} \right) \]

\[ \sum\limits_{j = 1}^n {j\left( {j + 1} \right)} \cdots \left( {j + k} \right) = \frac{1}{{k + 2}}\frac{{\left( {n + k + 1} \right)!}}{{\left( {n - 1} \right)!}} \]

\[ \sum\limits_{j = 1}^n {j\left( {j + 1} \right)^2 } = \frac{1}{{12}}n\left( {n + 1} \right)\left( {n + 2} \right)\left( {3n + 5} \right) \]

\[ \sum\limits_{j = 1}^n {j\left( {j + 1} \right)^2 } \left( {j + 2} \right) = \frac{1}{{10}}n\left( {n + 1} \right)\left( {n + 2} \right)\left( {n + 3} \right)\left( {2n + 3} \right) \]

\[ \sum\limits_{j = 1}^n {j\left( {n^2 - j^2 } \right)} = \frac{1}{4}n^2 \left( {n^2 - 1} \right) \]

\[ \sum\limits_{j = 1}^n {2^j j\left( {j + 1} \right) = 2^{n + 1} \left( {n^2 - n + 2} \right) - 4} \]

\[ \frac{1}{{1 \cdot 2}} + \frac{1}{{2 \cdot 3}} + \frac{1}{{3 \cdot 4}} + \cdots + \frac{1}{{n \cdot \left( {n + 1} \right)}} = 1 - \frac{1}{{n + 1}} = \frac{n}{{n + 1}} \]

\[ \frac{1}{{1 \cdot 2 \cdot 3}} + \frac{1}{{2 \cdot 3 \cdot 4}} + \frac{1}{{3 \cdot 4 \cdot 5}} + \cdots + \frac{1}{{n\left( {n + 1} \right)\left( {n + 2} \right)}} \] \[ = \frac{1}{4} - \frac{1}{{2\left( {n + 1} \right)\left( {n + 2} \right)}} \]

\[ \frac{1}{{1 \cdot 2 \cdot 3 \cdot 4}} + \frac{1}{{2 \cdot 3 \cdot 4 \cdot 5}} + \frac{1}{{3 \cdot 4 \cdot 5 \cdot 6}} + \cdots + \frac{1}{{n\left( {n + 1} \right)\left( {n + 2} \right)\left( {n + 3} \right)}} \] \[ = \frac{1}{{18}} - \frac{1}{{3\left( {n + 1} \right)\left( {n + 2} \right)\left( {n + 3} \right)}} \]

\[ \sum\limits_{j = 2}^n {\frac{1}{{\left( {j + 1} \right)\left( {j - 1} \right)}}} = \sum\limits_{j = 2}^n {\frac{1}{{j^2 - 1}} = \frac{3}{4} - \frac{1}{{2n}} - \frac{1}{{2\left( {n + 1} \right)}}} \]

\[ \sum\limits_{j = 1}^n {\frac{1}{{\left( {2j + 1} \right)\left( {2j - 1} \right)}}} = \frac{n}{{2n + 1}} \]

\[ \sum\limits_{j = 1}^n {\frac{1}{{\left( {3j - 2} \right)\left( {3j + 1} \right)}}} = \frac{n}{{3n + 1}} \]

\[ \sum\limits_{j = 1}^n {\frac{1}{{\left( {2j - 1} \right)\left( {2j + 1} \right)\left( {2j + 3} \right)}}} = \frac{1}{{12}} - \frac{1}{{4\left( {2n + 1} \right)\left( {2n + 3} \right)}} \]

\[ \sum\limits_{j = 1}^n {\frac{1}{{\left( {3j - 2} \right)\left( {3j + 1} \right)\left( {3j + 4} \right)}}} = \frac{1}{{24}} - \frac{1}{{6\left( {3n + 1} \right)\left( {3n + 4} \right)}} \]

\[ \sum\limits_{j = 1}^n {\frac{{2j - 1}}{{j\left( {j + 1} \right)\left( {j + 2} \right)}}} = \frac{3}{4} - \frac{2}{{n + 2}} + \frac{1}{{2\left( {n + 1} \right)\left( {n + 2} \right)}} \]

\[ \sum\limits_{j = 1}^n {\frac{{j + 2}}{{j\left( {j + 1} \right)\left( {j + 3} \right)}}} \] \[ = \frac{{29}}{{36}} - \frac{1}{{n + 3}} - \frac{3}{{2\left( {n + 2} \right)\left( {n + 3} \right)}} - \frac{4}{{3\left( {n + 1} \right)\left( {n + 2} \right)\left( {n + 3} \right)}} \]

\[ \sum\limits_{j = 1}^n {\frac{{j2^{j - 1} }}{{\left( {j + 1} \right)\left( {j + 2} \right)}}} = \frac{{2^n }}{{n + 2}} - \frac{1}{2} \]

\[ \sum\limits_{j = 1}^n {\frac{{j^2 4^j }}{{\left( {j + 1} \right)\left( {j + 2} \right)}}} = \frac{2}{3} + \frac{{\left( {n - 1} \right)4^{n + 1} }}{{3\left( {n + 2} \right)}} \]

\[ \sum\limits_{j = 1}^n {\frac{{j + 2}}{{j\left( {j + 1} \right)2^j }}} = 1 - \frac{1}{{\left( {n + 1} \right)2^n }} \]

\[ \sum\limits_{j = 1}^n {\frac{{2j + 3}}{{j\left( {j + 1} \right)3^j }}} = 1 - \frac{1}{{\left( {n + 1} \right)3^n }} \]

\[ \sum\limits_{j = 1}^n {\frac{{\left( { - 1} \right)^{j - 1} 2^j }}{{\left[ {2^j + \left( { - 1} \right)^j } \right]\left[ {2^{j + 1} + \left( { - 1} \right)^{j + 1} } \right]}}} = \frac{1}{3}\left[ {1 + \frac{{\left( { - 1} \right)^{n + 1} }}{{2^{n + 1} + \left( { - 1} \right)^{n + 1} }}} \right] \]

\[ \sum\limits_{j = 1}^n {\frac{{b\left( {b + 1} \right) \cdots \left( {b + j - 1} \right)}}{{a\left( {a + 1} \right) \cdots \left( {a + j - 1} \right)}}} = \frac{1}{{b - a + 1}}\left[ {\frac{{b\left( {b + 1} \right) \cdots \left( {b + n} \right)}}{{a\left( {a + 1} \right) \cdots \left( {a + n - 1} \right)}} - b} \right] \] 


参阅
  1. 数学符号
  2. 数学手册
  3. 初等数学
  4. 高等数学
  5. 公式图表
  6. 代数
  7. 几何
  8. 函数
  9. 例题
  10. 书单
  11. 数学软件 - 数学手册计算器


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